Lời giải:
Ta có:
\(\int ^{\frac{\pi}{2}}_{0}\frac{\sin x}{(\sin x+\cos x)^3}dx=\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\sin x}{(\sin x+\cos x)^3}dx+\int ^{\frac{\pi}{4}}_{0}\frac{\sin x}{(\sin x+\cos x)^3}dx\)
\(=A+B\)
Xét riêng rẽ:
\(A=\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\sin^3 x}{(\sin x+\cos x)^3}.\frac{dx}{\sin ^2x}=\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{1}{\left(\frac{\sin x+\cos x}{\sin x}\right)^3}d(-\cot x)\)
\(=\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{1}{(\cot x+1)^3}d(-\cot x)=-\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{d(\cot x+1)}{(\cot x+1)^3}\)
\(=\left.\begin{matrix} \frac{\pi}{2}\\ \frac{\pi}{4}\end{matrix}\right|\frac{1}{2(\cot x+1)^2}=\frac{3}{8}\)
\(B=\int ^{\frac{\pi}{4}}_{0}\frac{\sin x+\cos x-\cos x}{(\sin x+\cos x)^3}dx\)\(=\int ^{\frac{\pi}{4}}_{0}\frac{ 1}{(\sin x+\cos x)^2}dx-\int ^{\frac{\pi}{4}}_{0}\frac{\cos x}{(\sin x+\cos x)^3}dx\)
\(=\int ^{\frac{\pi}{4}}_{0}\frac{1}{\left(\frac{\sin x+\cos x}{\cos x}\right)^2}.\frac{dx}{\cos ^2x}-\int ^{\frac{\pi}{4}}_{0}\frac{1}{\left(\frac{\sin x+\cos x}{\cos^3 x}\right)^3}.\frac{dx}{\cos ^2x}\)
\(=\int ^{\frac{\pi}{4}}_{0}\frac{d(\tan x)}{(\tan x+1)^2}-\int ^{\frac{\pi}{4}}_{0}\frac{d(\tan x)}{(\tan x+1)^3}\)
\(=\int ^{\frac{\pi}{4}}_{0}\frac{d(\tan x+1)}{(\tan x+1)^2}-\int ^{\frac{\pi}{4}}_{0}\frac{d(\tan x+1)}{(\tan x+1)^3}\)
\(=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{-1}{\tan x+1}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{1}{2(\tan x+1)^2}=\frac{1}{8}\)
Do đó: \(\int ^{\frac{\pi}{2}}_{0}\frac{\sin x}{(\sin x+\cos x)^3}dx=\frac{3}{8}+\frac{1}{8}=\frac{1}{2}\)
Sở dĩ phải chia tích phân thành tổng nhỏ như vậy là do khi ta thực hiện chia sin x xuống dưới mẫu thì hàm số không liên tục trong đoạn \([\frac{\pi}{2}; 0]\)